Write (-3+4i)^(1+i) in the form u+iv (principal value)
Write $(-3+4i)^{1+i}$ in the form $u+iv$ (principal value).
Explanation of Method
Convert the base to polar form, compute its principal logarithm, multiply by the exponent $1+i$, then exponentiate back using $e^{a+ib}=e^a(\cos b+i\sin b)$.
Step 1 · Polar form of $-3+4i$
$$ r=|-3+4i|=\sqrt{9+16}=5 $$
Since $x=-3<0,\ y=4>0$ (second quadrant): $\ \theta=\pi-\arctan\frac43\approx\pi-0.9273\approx2.2143$.
Step 2 · Principal logarithm
$$ \text{Ln}(-3+4i)=\ln5+i\theta\approx1.6094+2.2143i $$
Step 3 · Multiply by $1+i$
$$ (1+i)(\ln5+i\theta)=(\ln5-\theta)+i(\ln5+\theta) $$
$$ a=\ln5-\theta\approx-0.6049, \qquad b=\ln5+\theta\approx3.8237 $$
Step 4 · Exponentiate
$$ (-3+4i)^{1+i}=e^{a+ib}=e^a(\cos b+i\sin b) $$
$$ e^a\approx0.54615, \qquad \cos b\approx-0.77622, \qquad \sin b\approx-0.63046 $$
Step 5 · Result
$$ \boxed{(-3+4i)^{1+i}=e^{\ln5-\theta}\left[\cos(\ln5+\theta)+i\sin(\ln5+\theta)\right]\approx-0.4239-0.3443i} $$
where $\theta=\pi-\arctan\frac43$.
Verification
Direct high-precision evaluation of $e^{(1+i)\,\text{Log}(-3+4i)}$ gives $-0.423935\ldots-0.344325\ldots i$, matching Step 5 to four decimal places.
Result
As with all non-integer complex powers, this is the principal branch — using $\theta+2k\pi$ in place of $\theta$ yields the full multi-valued family.